JUST THE MATHS SLIDES NUMBER 16.7 LAPLACE TRANSFORMS 7 (An - - PDF document

“JUST THE MATHS” SLIDES NUMBER 16.7 LAPLACE TRANSFORMS 7 (An appendix) by A.J.Hobson

One view of how Laplace Transforms might have arisen

UNIT 16.7 - LAPLACE TRANSFORMS 7 (AN APPENDIX) ONE VIEW OF HOW LAPLACE TRANSFORMS MIGHT HAVE ARISEN. (i) The problem is to solve a second order linear differ- ential equation with constant coefficients, ad2x dt2 + bdx dt + cx = f(t). (ii) We assume that the equivalent first order differential equation, adx dt + bx = f(t), has already been studied. We examine the following example: EXAMPLE Solve the differential equation, dx dt + 3x = e2t, given that x = 0 when t = 0.

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Solution The “integrating factor method” uses the coeffi- cient of x to find a function of t which multiplies both sides of the given differential equation to convert it to an “exact” differential equation The integrating factor in the current example is e3t, since the coefficient of x is 3. We obtain, e3t

  dx

dt + 3x

   = e5t.

This is equivalent to d dt

• xe3t
• = e5t.

Integrating both sides with respect to t, xe3t = e5t 5 + C

• r

x = e2t 5 + Ce−3t.

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Putting x = 0 and t = 0, we have 0 = 1 5 + C. Hence, C = −1

5, and the complete solution becomes

x = e2t 5 − e−3t 5 . (iii) We shall now examine a different way of setting out the above working in which the boundary condition is substituted earlier. We multiply both sides of the differential equation by e3t as before, then integrate both sides of the new “exact” equation from 0 to t.

t

d dt

• xe3t
• dt =

t

0 e5t dt.

That is,

• xe3t

t

0 =

   e5t

5

   

t

, giving xe3t − 0 = e5t 5 − 1 5 since x = 0 when t = 0.

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In other words, x = e2t 5 − e−3t 5 , as before (iv) We consider, next, whether an example of a sec-

• nd order linear differential equation could be solved by

a similar method. EXAMPLE Solve the differential equation, d2x dt2 − 10dx dt + 21x = e9t, given that x = 0 and dx

dt = 0 when t = 0.

Solution We assume that an integrating factor for this equation is est, where s, at present, is unknown, but is assumed to be positive Hence, we multiply throughout by est and integrate from 0 to t.

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t

0 est

   d2x

dt2 − 10dx dt + 21x

    dt = t

0 e(s+9)t dt

=

    

e(s+9)t s + 9

    

t

. Using “integration by parts”, and the boundary con- dition,

t

0 estdx

dt dt = estx − s

t

0 estx dt.

t

0 estd2x

dt2 dt = estdx dt − s

t

0 estdx

dt dt = estdx dt − sestx + s2 t

0 estx dt.

Substituting these results into the differential equation, we may collect together terms which involve

t

0 estx dt and est

as follows: (s2+10s+21)

t

0 estx dt+est

  dx

dt − (s + 10)x

   =     

e(s+9)t s + 9

    

t

.

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(v) OBSERVATIONS (a) If we had used e−st instead of est, the quadratic ex- pression in s, above, would have had the same coefficients as the original differential equation. That is, (s2 − 10s + 21). (b) Using e−st with s > 0, suppose we had integrated from 0 to ∞ instead of 0 to t. The term, est

  dx

dt − (s + 10)x

   ,

would have been absent, since e−∞ = 0. (vi) Having made our observations, we start again, mul- tiplying both sides of the differential equation by e−st and integrating from 0 to ∞. We obtain (s2 − 10s + 21)

∞

e−stx dt =

    

e(−s+9)t −s + 9

    

∞

.

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(s2 − 10s + 21)

∞

e−stx dt = −1 −s + 9 = 1 s − 9. Note: This works only if s > 9, but we can easily assume that it is so. Using s2 − 10s + 21 ≡ (s − 3)(s − 7),

∞

e−stx dt = 1 (s − 9)(s − 3)(s − 7). Using partial fractions,

∞

e−stx dt = 1 12. 1 s − 9 + 1 24. 1 s − 3 − 1 8. 1 s − 7. (vii) Finally, it can be shown, by an independent method

• f solution, that

x = e9t 12 + e3t 24 − e7t 8 . We may conclude that the solution of the differential equation is closely linked to the integral,

∞

e−stx dt, which is called the “Laplace Transform” of x(t).

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